Given two vectors `veca=-hati + 2hatj + 2hatk and vecb =- 2hati + hatj + 2hatk`
find `|vec a xx vec b|`

Explanation

To find the magnitude of the cross product of the two vectors \(\vec{a}\) and \(\vec{b}\), we can follow these steps:

### Step 1: Write down the vectors
Given:
\[ \vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k} \]
\[ \vec{b} = -2\hat{i} + \hat{j} + 2\hat{k} \]

### Step 2: Set up the determinant for the cross product
The cross product \(\vec{a} \times \vec{b}\) can be calculated using the determinant of a 3x3 matrix:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ -2 & 1 & 2 \end{vmatrix} \]

### Step 3: Calculate the determinant
To calculate the determinant, we expand it:
\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} -1 & 2 \\ -2 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} -1 & 2 \\ -2 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \(\hat{i}\):
\[ \begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix} = (2)(2) - (2)(1) = 4 - 2 = 2 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} -1 & 2 \\ -2 & 2 \end{vmatrix} = (-1)(2) - (2)(-2) = -2 + 4 = 2 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} -1 & 2 \\ -2 & 1 \end{vmatrix} = (-1)(1) - (2)(-2) = -1 + 4 = 3 \]

Putting it all together:
\[ \vec{a} \times \vec{b} = 2\hat{i} - 2\hat{j} + 3\hat{k} \]

### Step 4: Find the magnitude of the cross product
The magnitude of \(\vec{a} \times \vec{b}\) is given by:
\[ |\vec{a} \times \vec{b}| = \sqrt{(2)^2 + (-2)^2 + (3)^2} \]
Calculating this:
\[ |\vec{a} \times \vec{b}| = \sqrt{4 + 4 + 9} = \sqrt{17} \]

### Final Answer
Thus, the magnitude of the cross product \(|\vec{a} \times \vec{b}|\) is:
\[ \sqrt{17} \]

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